8.2. Python Lab: Systems of Equations#

Linear equation: If \(a\), \(b\), and \(c\) are real numbers, the graph of an equation of the form

\[\begin{align*} ax+by = c \end{align*}\]

is a straight line (if \(a\) and \(b\) are not both zero), so such an equation is called a linear equation in the variables \(x\) and \(y\).

Example: Solve

\[\begin{align*} \begin{cases} x+y=0\\ x-y=0,\\ \end{cases} \end{align*}\]

Graphically.

matplotlib:

# numpy
import numpy as np
# matplotlib
import matplotlib.pyplot as plt
%matplotlib inline
plt.style.use('seaborn-white')

x = np.linspace( -10 , 10 ,100)
fig, ax = plt.subplots(nrows=1, ncols=1, figsize=(6, 6), sharex = False)
_ = ax.plot(x, -x, color='r', label='x + y = 0')
_ = ax.plot(x, x, color='b', label='x - y = 0')
_ = ax.set_xlabel('$x$', fontsize=18)
_ = ax.set_ylabel('$y$', fontsize=18)
_ = ax.legend(loc='best', bbox_to_anchor=(1.01, 0., 0.2, 0.6))
_ = ax.set_xlim([min(x),max(x)])
_ = ax.set_ylim([min(x),max(x)])
_ = ax.grid(True, which='both')
_ = ax.axhline(y=0, color='k')
_ = ax.axvline(x=0, color='k')
../_images/PyLab_Systems_of_Equations_3_0.png

plotly:

## plotly
from plotly.offline import iplot 
import plotly.graph_objs as go

x = np.linspace( -10 , 10 ,100)
fig = go.Figure()
fig.add_trace(go.Scatter(x= x, y= -x, line=dict(color='Blue', width= 1.5), name = 'x + y = 0'))
fig.add_trace(go.Scatter(x= x, y= x, line=dict(color='OrangeRed', width= 1.5), name = 'x - y = 0'))
            
# Update
fig.update_layout(plot_bgcolor= 'white', width = 600)
fig.update_xaxes(showline=True, linewidth=1, linecolor='Lightgray', mirror=True,
                 showgrid=True, gridwidth=1, gridcolor='Lightgray',
                 zeroline=True, zerolinewidth=1.5, zerolinecolor='black')
fig.update_yaxes(showline=True, linewidth=1, linecolor='Lightgray', mirror=True,
                 showgrid=True, gridwidth=1, gridcolor='Lightgray',
                 zeroline=True, zerolinewidth=1.5, zerolinecolor='black')
fig.update_xaxes(title_text='x', range=[-10, 10])
fig.update_yaxes(title_text='y', range=[-10, 10])
fig.show()

Example: Solve

\[\begin{align*} \begin{cases} x+y+z=2\\ x+y+z=5 \end{cases}. \end{align*}\]

matplotlib:

fig = plt.figure(figsize = (20, 10))
ax = fig.add_subplot(111, projection='3d')

# Make data.
X = np.linspace( -10 , 10 ,100)
Y = np.linspace( -10 , 10 ,100)
X, Y = np.meshgrid(X, Y)
Z1 = 2 - X - Y 
Z2 = 5 - X - Y 

# Plot the surface.
surf = ax.plot_surface(X, Y, Z1, color = 'blue', linewidth=0, antialiased=False)
_ = ax.plot_surface(X, Y, Z2, color = 'lightgreen', linewidth=0, antialiased=False)
_ = ax.set_zlim(-20, 40)
# _ = ax.zaxis.set_major_locator(LinearLocator(10))
_ = ax.set_xlabel('x', fontsize=18)
_ = ax.set_ylabel('y', fontsize=18)
_ = ax.set_zlabel('z', fontsize=18)
../_images/PyLab_Systems_of_Equations_8_0.png

plotly:

X = np.linspace( -10 , 10 ,100)
Y = np.linspace( -10 , 10 ,100)
X, Y = np.meshgrid(X, Y)
Z1 = 2 - X - Y 
Z2 = 5 - X - Y 

fig = go.Figure()
fig.add_trace(go.Surface(z=Z1, x=X, y=Y, colorscale='Greens', colorbar_len=0.5))
fig.add_trace(go.Surface(z=Z2, x=X, y=Y, colorscale='Blues', colorbar_len=0.5))

fig.update_layout(scene = dict(xaxis = dict(tickvals= list(np.arange(-10,11,5))),
                               yaxis = dict(tickvals= list(np.arange(-10,11,5))),
                               zaxis = dict(nticks=5, ticks='outside',tick0=0, tickwidth=4),),
                  margin=dict(r=10, l=10, b=10, t=10))
fig.show()

8.2.1. Solving linear system using back-substitution#

Example: Solve the following linear system using back-substitution.

\[\begin{equation*} \begin{cases} x_{1}+2\,x_{2}+3\,x_{3}=2\\ 4\,x_{1}-3\,x_{2}+x_{3}=-3\\ x_{2}-2\,x_{1}+3\,x_{3}=5 \end{cases} \end{equation*}\]

The corresponding augmented matrix is

from __future__ import division
from sympy import init_session, init_printing, Matrix
init_printing(use_unicode=True)

M = Matrix([[1 , 2 , 3 , 2], [4 , -3 , 1 , -3], [-2 , 1 , 3 , 5]]) 
M
../_images/PyLab_Systems_of_Equations_14_0.png

In RREF:

RREF= M.rref()[0]
RREF
../_images/PyLab_Systems_of_Equations_16_0.png

Therefore, the solutions are:

from IPython.display import display
display(r'x_1 = %i, x_2 = %i and x_3 = %i' % (RREF[0,-1], RREF[1,-1], RREF[2,-1]))

'x_1 = -1, x_2 = 0 and x_3 = 1'

To verify our answer, we can use linalg from Numpy. This requires converting the Matrix M into an array.

A = np.asarray(M[:,:-1], dtype = int)
B = np.asarray(M[:,-1], dtype = int)
Matrix(np.linalg.solve(A, B))
../_images/PyLab_Systems_of_Equations_20_0.png